Finite-dimensional Hilbert $C^*$-modules

As a vector space, okay. Let $A$ be a C*-algebra and $E$ a Hilbert module over $A$ which is finite dimensional as a vector space. Then $A$ acts by left multiplication on $E$, i.e., we have a bounded homomorphism from $A$ into $B(E) \cong M_n$. Letting $I$ be the kernel of this homomorphism, this shows that $A$ has a finite codimension closed ideal $I$ (which is automatically self-adjoint) and $E$ is a Hilbert module over $A/I$ (exercise).

So we have to classify finite dimensional Hilbert modules over finite dimensional C*-algebras. Any finite dimensional C*-algebra is a direct sum of matrix algebras, and any Hilbert module over such a direct sum decomposes as a direct sum of Hilbert modules over the summand matrix algebras. So we only have to classify the finite dimensional Hilbert modules over $M_n$. But again, left multiplication gives a homomorphism from $M_n$ into $B(E)$, and this $*$-homomorphism must be unital (exercise), which by the answer to this question means that $E \cong \mathbb{C}^n\otimes \mathbb{C}^k$ for some $k$, with the action being $a(v\otimes w) = av\otimes w$. So $E$ is a direct sum of finitely many copies of $\mathbb{C}^n$ as a module over $M_n$ in the natural way. The inner product is described in Matthew's answer.

Let $E$ be a Hilbert $C^*$-module over a $C^\ast$-algebra $A$. (I'll use the right module convention, which basically just makes life difficult for myself.) Then $I$, the closure of $\newcommand{\lin}{\operatorname{lin}} \lin \{ (x|y):x,t\in E \}$ is a (two-sided) closed ideal in $A$. (Because, for example $(x|y)a = (x|y\cdot a)\in I$). Recall that $E$ is full when $A=I$.

Now suppose that $E$ is a finite-dimensional vector space. Then $\lin \{ (x|y):x,y\in E \}$ is finite-dimensional, so equals $I$. As $I$ is itself a $C^\ast$-algebra, and is finite-dimensional, it in particular has a unit $e$. Then $I = eA = Ae = AeA$ and $A\rightarrow I, a\mapsto ae = eae = ea$ is a $*$-homomorphism which is the identity on $I$. It follows that $A \cong I \oplus B$ for some $C^\ast$-algebra $B$.

We know that $E = E\cdot I$ and so $x = x\cdot e$ for each $x\in E$. We can hence treat $E$ is a full Hilbert $C^\ast$-module over $I$. We now use that $I$ is a direct sum of matrix algebras, say $I = \sum_{k=1}^N M_{n_k}$. Let $1_k$ be the unit of $M_{n_k}$ so $x = x\cdot e = x\cdot(\sum 1_k) = \sum_k x\cdot 1_k$ for $x\in E$, and so $E = \sum_{k=1}^N E_k$ where $E_k = E\cdot 1_k$ is a full module over $M_{n_k}$.

Finally, we classify modules $F$ over $M_n$. One can show (I do not know a fast way to show this) that if $F$ is a finite-dimensional vector space with a (left) action of $M_n$ then $F \cong \mathbb C^n \otimes F'$ where $M_n$ acts on $\mathbb C^n$ in the usual way (treat $\mathbb C^n$ as column vectors) and the action on $F$ is as $a\otimes 1$. As $F$ is a right module over $M_n$, we have that $(\xi\otimes x')\cdot a = a^\top\xi \otimes x'$ where $a^\top$ is the transpose of $a\in M_n$. Let $\delta=\delta_1$ be the first unit vector in the standard basis of $\mathbb C^n$, so $a^\top\delta\otimes x'$ is a typical elementary vector in $F$.

Now consider the $M_n$-valued inner-product on $F$ which is $$ (a^\top\delta\otimes x'|b^\top\delta\otimes y') = a^\ast (\delta\otimes x'| \delta\otimes y') b = a^\ast \theta(x'|y')b $$ say, for some sesquilinear $\theta:F\times F\rightarrow M_n$. However, notice that if $a^\top\delta=0$ or $b^\top\delta=0$ then $a^*\theta(x'|y')b=0$ for any $x',y'$. Let $\theta_{\xi,\eta}\in M_n$ be the rank-one operator $\alpha\mapsto \xi(\eta|\alpha)$. Further let $J:\mathbb C^n\rightarrow \mathbb C^n$ be the anti-linear operator given by pointwise complex conjugation. Then $a^\top = Ja^\ast J$. Thus, if $(\xi|\delta)=0$ or $(\eta|\delta)=0$, then $\theta_{\xi,\alpha}^\ast \theta(x',y') \theta_{\eta,\beta}=0$ for any $\alpha,\beta$, so $(\xi|\theta(x',y')\eta)=0$. It follows that $\theta(x',y') \in \mathbb C \theta_{\delta,\delta}$. So, there is a Hilbert space structure on $F'$ with $\theta(x',y') = (x'|y') \theta_{\delta,\delta}$. We conclude that the $M_n$-valued inner-product on $F=\mathbb C^n\otimes F'$ is $$ (\xi\otimes x'|\eta\otimes y') = (x'|y') \theta_{\eta,\xi} \in M_n. $$

So, this completely answers the case of a finite-dimensional module over $A$. There was also the question of countable direct sums of such things. I'm not sure what is meant here. If you know that $E$ is a countable direct sum $E\cong \oplus E_n$ each $E_n$ finite-dimensional, then you can piece together the structure for each $E_n$. However, if you want to know which $E$ can occur in this way, then that seems more complicated.